TRANSIENT RESPONSE OF COMPOSITE AXISYMMETRIC SHELLS SUBMITTED TO PYROTECHNIC SHOCKS Application to a payload adapter

A. Legay and J.-F. De¨ u Structural Mechanics and Coupled Systems Laboratory - Cnam Paris

30 july 2008 - WCCM8 - ECCOMAS 2008

TRANSIENT RESPONSE OF COMPOSITE AXISYMMETRIC SHELLS SUBMITTED TO PYROTECHNIC SHOCKS, Application to a

Application to a payload adapter

last separation phase: pyrotechnic cord explosion induces wave propagations in the adapter susceptible to damage the satellite → parametric analyses to design this problem 2 / 18

Motivation for a Fourier series model A pyrotechnic shock induces high peak acceleration and high-frequency content: a tridimensional analysis requests a huge number of d.o.f. and an extremely fine time discretization

→ take benefit of the geometry axisymmetry through a Fourier circumferential decomposition classical for static, modal and buckling analyses since the pionner work of E.L. Wilson (1965) still original for transient responses even more original for complex shock loadings like moving load due to pyrotechnic cord

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Kinematics of the axisymmetric geometry Tronconic shell element Displacements

Uz ~2 E

Ux γ

x

Ux

= u(x, θ) + zβ(x, θ)

Uθ

= v (x, θ) + zγ(x, θ)

Uz

= w (x, θ)

Uθ θ

R(x)

β ~1 E

z: distance between point and shell mean surface Kirchhoff-Love hypothesis

~3 E α

β

=

γ

=

∂w ∂x cos α 1 ∂w v− R R ∂θ

−

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Fourier Decomposition Displacements (•S for symmetric and •A for anti-symmetric) u(x, θ) = u0 (x) +

P∞

unS cos nθ +

v (x, θ) = v0 (x) +

P∞

vnS sin nθ +

n=1

w (x, θ) = w0 (x) +

n=1

P∞

n=1

P∞

unA sin nθ

P∞

vnA cos nθ

wnS cos nθ +

n=1

n=1

P∞

n=1

wnA sin nθ

Strains (Membrane and curvature splitting) 2

3 2 3 κxx xx ε = 4 θθ 5 + z 4 κθθ 5 =  + zκ 2xθ κxθ =

P∞

cos nθ +

κ=

P∞

κSn cos nθ +

S n=1 n

n=1

P∞

A n=1 n

P∞

n=1

sin nθ

κAn sin nθ

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Laminated composite material Payload adapter: alu. honeycomb core + carbon laminated faces ~ ez N k

zN

zk zk−1

~ ex z0

1

By denoting Qk stiffness matrix for layer k: I0 =

N X

ρk (zk − zk−1 ); I1 =

k=1

Aij =

N X k=1

Qkij (zk − zk−1 ); Bij =

N N 1X 1X 2 3 ρk (zk2 − zk−1 ); I2 = ρk (zk3 − zk−1 ) 2 k=1 3 k=1 N N 1X k 2 1X k 3 2 3 Qij (zk − zk−1 Q (z − zk−1 ); Dij = ) 2 k=1 3 k=1 ij k

Membrane-bending coupling for isotropic material: 0 −1 −Eh3 cos α @ ρh3 cos α 0 ; B= I1 = 2 12 R 12R(1 − ν ) 0

0 1 0

1 0 0 A

1−ν 2

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Finite element discretization Elemental nodal unknowns q en = [ un1 vn1 wn1 βn1 un2 vn2 wn2 βn2 ]T Mixed linear and cubic interpolation   un  vn  = Nm (x)q e and n wn Discretized gradient operators:   un nS or A = Dm  vn  wn   un κnS or A = Db  vn  wn

 βne  γne  = Nb (x)q e n 0 

⇒

Bm = D m N m

⇒

Bb = D b N b

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Discretized transient problem Elemental mass and Stiffness matrices Z le  T  T T Men = π(1+δ0n ) I0 Nm Nm + I1 (NT b Nm + Nm Nb ) + I2 Nb Nb R(x)dx 0

Ken = π(1+δ0n )

Z

le

 T  T T Bm ABm + BT m BBb + Bb BBm + Bb DBb R(x)dx

0

Global system to be solved (small size because of 1D mesh) For harmonic n = 0 Mn q¨0 + Kn q 0

=

F0

For each harmonic n > 0 Mn q¨n S + Kn q n S

=

FSn for symmetric part

Mn q¨n A + Kn q n A

=

FAn for anti-symmetric part

+ initial conditions

Resolution by implicit Newmark time scheme, γ = 21 , β =

1 4 8 / 18

Non-axisymmetric normal load Description of load: θ

θ

lineic load p for θ ∈ [θref − 2p ; θref + 2p ] and a fixed radius Rp ; total force P = pRp θc for θp → 0, equivalent to concentrated load

Virtual external work: symmetric part of harmonic n, lineic load

E~1

P = rp θp

δW

p

E~2

Z

θref +

= θref −

rp θref

S

θp

= 2p

θp 2

θp 2

p cos (nθ)rp dθ δwnS

rp θp sin (n ) cos (nθref ) δwnS n 2

If θp ≈ 0 then δW S = P cos (nθref ) δwnS 9 / 18

External load vector Fixed load in space: P cos (nθref )

for symmetric part

P sin (nθref )

for anti-symmetric part

Rotating load, constant angular speed: θref (t) = ωt P cos (nωt)

for symmetric part

P sin (nωt)

for anti-symmetric part

Note that for n = 0 only symmetric part has to be used More complex load can be investigated from virtual external work

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Validation: circular plate, concentrated static load Comparison with a 3D shell Nastran computation Fourier Nastran

1 2 3 5 Nastran 10 Normal displacement vs radius

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Circular plate, harmonic contributions

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Application to payload adapter h = 4.1mm, aluminium h = 11.8mm, composite, 3 layers h = 4.1mm, aluminium h = 6mm, aluminium h = 15mm, aluminium θp

1.92m diameter P

Two different loads: Rotating load:

Localized load:

θp = ωt, speed 7100m.s−1

P 4000N time

θp = 0, P = P(t)

0.02ms

1 round=0.85ms P = 4000N constant 13 / 18

Payload: transient response of a localized load Comparison with a 3D shell Nastran computation (implicit), harmonic contributions

Fourier, 1ms

Nastran, 1ms

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Payload: transient response of a rotating load Fourier 3D visualisation

0.025ms

0.25ms

0.5ms

0.75ms

1.00ms

1.25ms

1.50ms

1.25ms

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Payload: transient response of a rotating load Influence of number of harmonics

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Payload: transient response of a rotating load Comparison with Nastran (implicit) and Ls-Dyna (explicit) Mesh for Nastran/Ls-Dyna (46400 elements)

Computational time on a 2.8Ghz processor Nastran: 6h00 Ls-Dyna: 2h00 Fourier: 30s by harmonic, 10min for 20 harmonics ⇒ factor from 10 to 30 17 / 18

Conclusions Key words axisymmetric geometric Fourier serie decomposition complex load linear analysis

Difficulties pre-processor: Fourier decomposition of the load, ... processor: automatization of the loop over harmonics post-processor: 3D visualization, ...

Advantages choice of a physical harmonic basis; only few harmonics are needed even for wave propagation problem decoupling of the harmonics ⇒ natural parallel computation with no further developments computational time divided by 10 to 30 (with 1 processor): very interesting for parametric studies

Thanks to Lionel Auffray for Nastran and Ls-Dyna computations 18 / 18